# I Help with Euler-Lagrange Equation

1. Jun 25, 2016

### Markus Hanke

I have begun teaching myself Lagrangian field theory in preparation for taking the plunge into quantum field theory ( it's just a hobby, not any kind of formal course ). When working through exercises, I have run across the following issue which I don't quite understand. I am being given a Lagrangian density, and ask to derive the equations of motion; I understand the principles involved, and everything is fine and easy until I get to the point where I need to evaluate the following expression :

$$\frac{1}{2}\frac{\partial }{\partial \left ( \partial _{\mu}\varphi \right )}\left ( \partial _{\mu}\varphi \right )^2$$

wherein $\varphi(x,y,z,t)$ is a scalar field. My approach to this was as simple as it was naive :

$$\frac{1}{2}\frac{\partial }{\partial \left ( \partial _{\mu}\varphi \right )}\left ( \partial _{\mu}\varphi \right )^2=\frac{1}{2}\frac{\partial }{\partial \left ( \partial _{\mu}\varphi \right )}\left ( \partial _{\mu}\varphi \right )\left ( \partial _{\mu}\varphi \right )$$

which evaluates to $\partial _{\mu}\varphi$ via the product rule. However, this is where I get stuck, because the answer is wrong - the correct approach should have been

$$\frac{1}{2}\frac{\partial }{\partial \left ( \partial _{\mu}\varphi \right )}\left ( \partial _{\mu}\varphi \right )\left ( \partial ^{\mu}\varphi \right )$$

which apparently evaluates to $\partial ^{\mu}\varphi$ ( though I have difficulties with that as well, but that's a separate issue ), and leads to the correct equations of motion. My question is : why is $\left ( \partial _{\mu}\varphi \right )^2=\left ( \partial _{\mu}\varphi \right )\left ( \partial ^{\mu}\varphi \right )$ and not $\left ( \partial _{\mu}\varphi \right )^2=\left ( \partial _{\mu}\varphi \right )\left ( \partial _{\mu}\varphi \right )$ ? I know that this is probably something very elementary, so please don't laugh at me, but I genuinely don't get it.

2. Jun 26, 2016

### stevendaryl

Staff Emeritus
This is not specific to quantum field theory, but is a fact about vectors and covectors.

A vector is something that is generally written with upper-indices. For example, if a massive particle's position is $x^\mu$, then its 4-velocity is given by $\frac{dx^\mu}{d\tau}$ (where $\tau$ is proper time for the particle), which is a vector. In contrast, a covector is written using lower-indices. For example, if $\phi$ is a scalar field, then its "gradient" is $\partial_\mu \phi \equiv \frac{\partial \phi}{\partial x^\mu}$.

To make a scalar, you have to use a vector together with a covector. That means that you have to have one object with upper-indices and another object with lower-indices. So if $A^\mu$ is a vector and $B_\mu$ is a covector, then you can make a scalar by combining them via: $\sum_\mu A^\mu B_\mu$. (It's usually written without the $\sum$, using the convention that repeated indices, one upper and one lower, are always summed over.) For example, we can combine a 4-velocity $\frac{dx^\mu}{d\tau}$ and a covector $\partial_\mu \phi$ to get the combination $\frac{dx^\mu}{d\tau} \partial_\mu \phi$. This is a scalar, the rate of change of $\phi$ for the particle as a function of proper time.

So if you have to use a vector and a covector to make a scalar, then how do you take the square of a vector? The answer is that you can't take the square of a vector without help from a metric tensor. The metric tensor $g_{\mu \nu}$ is an operator that converts a vector into a covector: If you have a vector $A^\mu$ then you can convert it into a corresponding covector $A_\mu[itex] through [itex]A_\mu = g_{\mu \nu} A^\nu$ (by convention, the repeated index $\nu$ on the right side of = is summed over). Using the metric, we can square a vector as follows:

$|A^\mu|^2 \equiv A^\mu A_\mu \equiv g_{\mu \nu} A^\mu A^\nu$ (in the last expression, both $\mu$ and $\nu$ are summed over).

Similarly, the square of a covector must involve a metric, as well:
$|\partial_\mu \phi|^2 \equiv \partial^\mu \phi \partial_\mu \phi \equiv g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi$
(where $g^{\mu \nu}$ is the inverse of $g_{\mu \nu}$; it converts a covector $\partial_\mu \phi$ into a vector, $\partial^\mu \phi$).

In quantum field theory, if an expression like $(\partial_\mu \phi)^2$ appears in the Lagrangian, it always means $\partial^\mu \phi \partial_\mu \phi$, because you can't take the square of a covector, otherwise.

3. Jun 26, 2016

### stevendaryl

Staff Emeritus
Oh, one further point. Since $\partial^\mu \phi \partial_\mu \phi$ means $g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi$, there are two factors of $\partial_\mu \phi$, so taking the derivative with respect to $\partial_\mu \phi$ gives a factor of two.

4. Jun 26, 2016

### Markus Hanke

My goodness, of course !! I had come across that already in my studies of GR, but it had completely slipped my mind in this context. I knew I was missing something elementary. Thank you for taking the time to reply in such detail, I understand it now

5. Jun 26, 2016

### Markus Hanke

Yup indeed :) Given the correct expansion of this bracket, I actually got the rest of the exercise done with no problems, it was really just this one upper index that tripped me up. But all is good now